Let \(P\) be the operator on \(\RR^2\) which projects each vector onto the \(x\)-axis, parallel to the \(y\)-axis: \(P(x, y) = (x, 0)\). Show that \(P\) is linear. What is the minimal polynomial for \(P\)?
Solution
For \((x_1,y_1),(x_2,y_2)\in\mathbb R^2\) and \(a,b\in\mathbb R\), \[ P\bigl(a(x_1,y_1)+b(x_2,y_2)\bigr) =(ax_1+bx_2,0) =aP(x_1,y_1)+bP(x_2,y_2), \] so \(P\) is linear.
Also \[ P^2=P, \] so \(P(P-I)=0\), and therefore the minimal polynomial divides \(x(x-1)\). Since \(P\ne0\) and \(P\ne I\), neither \(x\) nor \(x-1\) annihilates \(P\). Thus \[ m_P(x)=x(x-1). \]